rpnom
Badges: 2
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 6 years ago
#1
Name:  function.png
Views: 114
Size:  13.1 KB

I can do (i) but not (ii). The answer is h^-1(x)=h(x) but I just don't understand it! Thanks!
0
reply
Math12345
Badges: 14
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 6 years ago
#2
hh(x)=x

Multiply on the left by h^{-1}

h^{-1}hh(x)=h^{-1}(x)

So

h(x)=h^{-1}(x) since h^{-1}h=1 (identity)
0
reply
Zacken
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 6 years ago
#3
(Original post by rpnom)

I can do (i) but not (ii). The answer is h^-1(x)=h(x) but I just don't understand it! Thanks!
Just some alternative wording to the above:

We have h(h(x)) = x. Now you can apply a function to both sides of an equality (if it is injective, for the rigorous) so, for example when you have 2 = 2 you can apply the square function to both sides of the equality to get 2^2 = 2^2.

Now you have h(h(x)) = x. If we apply the function h^{-1}(x) (that is, the inverse function h) then we get h^{-1}(h(h(x))) = h^{-1}(x).

But when we apply the inverse function to the function itself, they just cancel out. So for example, when we do the square root of a square, it cancels out. (\sqrt{x})^2 = x. They just cancel each other out, you see? Because squaring and square rooting are inverses of one another.

So when we apply h^{-1}h(\text{anything}) we get \text{anything} back out.

In this case, our anything is h(x). So we have h^{-1}h(h(x)) = h(x) since the h^-1 cancels the first h.

So this means that h^{-1}(h(h(x))) = h^{-1}(x) is really just h(x) = h^{-1}(x).

Let's recap:

We had h(h(x)) = x.

We apply the inverse function h^{-1} to both sides.

We get h^{-1}(h(h(x))) = h^{-1}(x).

We simplify the LHS by noting that the inverse function of a function cancels out.

We get h(x) = h^{-1}(x).
1
reply
rpnom
Badges: 2
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report Thread starter 6 years ago
#4
(Original post by Zacken)
Just some alternative wording to the above:

We have h(h(x)) = x. Now you can apply a function to both sides of an equality (if it is injective, for the rigorous) so, for example when you have 2 = 2 you can apply the square function to both sides of the equality to get 2^2 = 2^2.

Now you have h(h(x)) = x. If we apply the function h^{-1}(x) (that is, the inverse function h) then we get h^{-1}(h(h(x))) = h^{-1}(x).

But when we apply the inverse function to the function itself, they just cancel out. So for example, when we do the square root of a square, it cancels out. (\sqrt{x})^2 = x. They just cancel each other out, you see? Because squaring and square rooting are inverses of one another.

So when we apply h^{-1}h(\text{anything}) we get \text{anything} back out.

In this case, our anything is h(x). So we have h^{-1}h(h(x)) = h(x) since the h^-1 cancels the first h.

So this means that h^{-1}(h(h(x))) = h^{-1}(x) is really just h(x) = h^{-1}(x).

Let's recap:

We had h(h(x)) = x.

We apply the inverse function h^{-1} to both sides.

We get h^{-1}(h(h(x))) = h^{-1}(x).

We simplify the LHS by noting that the inverse function of a function cancels out.

We get h(x) = h^{-1}(x).
Thank you!!!!
0
reply
Zacken
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report 6 years ago
#5
(Original post by rpnom)
Thank you!!!!
No worries.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Year 12s - where are you at with making decisions about university?

I’ve chosen my course and my university (18)
31.03%
I’ve chosen my course and shortlisted some universities (22)
37.93%
I’ve chosen my course, but not any universities (2)
3.45%
I’ve chosen my university, but not my course (3)
5.17%
I’ve shortlisted some universities, but not my course (4)
6.9%
I’m starting to consider my university options (7)
12.07%
I haven’t started thinking about university yet (1)
1.72%
I’m not planning on going to university (1)
1.72%

Watched Threads

View All